Understanding the Intrinsic Parity of Quarks and Antiquarks

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A quark and antiquark have opposite parity. The quark is customarily taken to have positive parity. I understand this to mean that Pf = f, where f is the wavefunction of the quark and Pg = -g, where g is the wavefunction of the antiquark.

Does this mean that P acting on an antiquark wavefunction flips the handedness of the particle, but won't do this for quarks? I don't think that makes sense.
 
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Doesn't it flip the handedness of both? The space parity operator for a Dirac particle is γ0.

I'm not sure which property you mean by handedness, could be either helicity or chirality. The chirality operator is γ5, which anticommutes with γ0, hence changes sign under parity. The helicity operator is ∑·p, and ∑ commutes with γ0 but p changes sign, hence the helicity changes sign under parity also.

The intrinsic parity of a fermion is not uniquely defined, there's a choice of phase involved. The relative parity of two different fermions is well defined, and as you said, the antiparticle and the particle have opposite parity.
 
Bill_K said:
Doesn't it flip the handedness of both? The space parity operator for a Dirac particle is γ0.

I'm not sure which property you mean by handedness, could be either helicity or chirality. The chirality operator is γ5, which anticommutes with γ0, hence changes sign under parity. The helicity operator is ∑·p, and ∑ commutes with γ0 but p changes sign, hence the helicity changes sign under parity also.

The intrinsic parity of a fermion is not uniquely defined, there's a choice of phase involved. The relative parity of two different fermions is well defined, and as you said, the antiparticle and the particle have opposite parity.

Thanks for the reply! If both chirality and helicity are flipped for both particle and antiparticle, then what's the consequence of them having opposite relative parity? Doesn't the +1 eigenvalue case mean that the function is left unchanged under the parity operation?
 
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